anonymous
  • anonymous
Solve these inequalities. (ii) 3^x +5 bigger or equal to 32 Please help! I don't know how to solve this though I a aware it involves logarithms. Please show me what to do and explain why.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nnesha
  • Nnesha
\[\huge\rm 3^x +5 \ge 32\] yes right you have to take log both sides but first move `5` to the right side and then take log
Nnesha
  • Nnesha
make sense ?
anonymous
  • anonymous
So It would be \[3^{x} \ge 17\] ? How do I take log of both sides?

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Nnesha
  • Nnesha
17 ?? i guess thats a typo
anonymous
  • anonymous
Um no actually. What did I do wrong?
Nnesha
  • Nnesha
how did you get 17 ?
anonymous
  • anonymous
I subtracted 5 from both sides?
Nnesha
  • Nnesha
yes right so 32-5 isn't equal to 17 :=)
anonymous
  • anonymous
Ooooh. I do mistakes like that often. Oops!
anonymous
  • anonymous
ok so after \[3^{x} \ge 27\] how do I take log of both sides?
Nnesha
  • Nnesha
yes right now take log both sides and apply the product rule \[\huge\rm log (3)^x \ge \log(27)\]
Nnesha
  • Nnesha
power rule ***
anonymous
  • anonymous
Ok I am new too logarithms so sorry for taking so much time. If you use the power rule does that mean that I move the exponent x in front of the first logarithm?
Nnesha
  • Nnesha
yes right :=) and its okay take ur time!
anonymous
  • anonymous
Thanks! All right so, \[x \log_{}3 \ge \log27\]
Nnesha
  • Nnesha
yes right now to solve for x you should move log(3) to the right side :=) and then use calculator :P that's it!
anonymous
  • anonymous
Ok, thank you so much! :D
Nnesha
  • Nnesha
what did you get ?? o.O
anonymous
  • anonymous
x\[x \ge 3\]
Nnesha
  • Nnesha
yes right!
Nnesha
  • Nnesha
good job!
anonymous
  • anonymous
Hooray!

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